Question

Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C. Round your answer to the nearest hundredth.

Answer 1

Given :

A constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C.

To Find :

The pressure change.

Solution :

By Gay-Lussac's Law :

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\\dfrac{1}{30+273}=\dfrac{P_2}{40+273}\\\\P=\dfrac{313}{303}\\\\P=1.033\ atm$

Change in pressure,

$\Delta \ P=P_2-P_1\\\\\Delta \ P=(1.033-1)\ atm\\\\\Delta \ P=0.33\ atm$

Therefore, the pressure change is 0.33 atm.

Hence, this is the required solution.